63. Unique Paths II java solutions

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

该题和

Unique Paths java solutions

相比就是多了个障碍设置，只要dp初始化对障碍特殊设置一下，在中间dp 的过程加下判断即可。

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) return 0;
 4         int m = obstacleGrid.length, n = obstacleGrid[0].length;
 5         int[][] dp = new int[m][n];
 6         boolean flag = true;
 7         for(int i = 0; i < m; i++){
 8             if(obstacleGrid[i][0] == 0 && flag == true){
 9                 dp[i][0] = 1;
10             }else{
11                 flag = false;
12                 dp[i][0] = 0;
13             }
14         }
15         flag = true;
16         for(int i = 0; i < n; i++){
17             if(obstacleGrid[0][i] == 0 && flag == true){
18                 dp[0][i] = 1;
19             }else{
20                 flag = false;
21                 dp[0][i] = 0;
22             }
23         }
24         for(int i = 1; i < m; i++){
25             for(int j = 1; j < n; j++){
26                 if(obstacleGrid[i][j] == 0)
27                     dp[i][j] = dp[i-1][j] + dp[i][j-1];
28                 else
29                     dp[i][j] = 0;
30             }
31         }
32         return dp[m-1][n-1];
33     }
34 }

 优化为一维数组解法：

 1 public class Solution {
 2     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
 3         if(obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length == 0) return 0;
 4         int[] dp = new int[obstacleGrid[0].length];
 5         dp[0] = 1;
 6         for (int i = 0; i < obstacleGrid.length; i++)
 7             for (int j = 0; j < obstacleGrid[0].length; j++)
 8                 if (obstacleGrid[i][j] == 1) dp[j] = 0;
 9                 else if (j > 0) dp[j] += dp[j - 1];
10         return dp[obstacleGrid[0].length - 1];
11     }
12 }

            dp[j]  上一行，

dp[j-1]           当前位置的左边

 

因此可以优化为一维数组。

posted @ 2016-07-03 11:16  Miller1991  阅读(162)  评论(0)  编辑  收藏  举报